Question

7. Consider the two functions: f(x, a)-40 - 3xa 9(2) 10 + 7x (a) Find the x value of the point where the two equa- (b) Find the value of the functions at the point where erwya tions intersect (in terms of the variable a) the two equations intersect (in terms of the variable a) (c) Take the partial derivate of f with respect to , and with respec t to a. d) What are the values of these derivatives when r - 3 2, which can be written as and a (3.2) and 3, 2 (e) Next, calculate these two numbers v1 = (3.2) yatt er v2 = Con (3.2) T(3,2)
E and F Specifically Please

Answer 1

Ans (a) It has been given in the question that:-

f(x,a) = 40-3xa , and

g(x) = 10+7x

In order to find out the x value of the point where the two equations intersect ( in terms of the variable a ), let us equate the equations. So , f(x,a)=g(x)

40-3xa= 10+7x

40-10= 7x+3xa

30=7x+3xa

So , it comes out as 30 = x(7+3a)

So, x= 30/7+3a

Ans (b) The value of the functions at the point where the two equations intersect (in terms of the variable a ) can be calculated in the following manner:-

Let us put the value of x in f(x, a) in order to find out the point where the two equations intersect.

f(x,a) = 40-3xa

= 40-(3*30*a/7+3a)

= 40-90a/7+3a

= 280+120a-90a/7+3a

= (280+30a)/7+3a

Ans(c) The partial dervative $\small \frac{\partial }{\partial x}$ can be calculated in the following manner:-

  $\small \frac{\partial }{\partial x}(10+7x)$ = $\small \frac{\partial }{\partial x}(10)+\frac{\partial }{\partial x}(7x)$

= 0+7=7

However, the partial derivative .$\small \int_{a}^{f}$ can be calculated in the following manner:-

$\small \int_{a}^{f}$40 -$\small \int_{a}^{f}$3xa

= 0-3x= -3x

Ans (d) However, the values of these derivatives when x=3 and a =2 can be calculated in the following manner:-

The value of .$\small \frac{\partial }{\partial x}$ is a constant number i.e. 7. So, it is not possible to calculate its value further and at each value of x, its value will remain 7 only.

A derivative represents the rate of change at any instant of data in the space your function covers. A constant is a constant and a function that equals a constant will never change, regardless of what data you give it. So, that is why its derivative is also zero because its value never changes.

However, the value of the derivative $\small \int_{a}^{f}$ was calculated in part (c) and its value came out as -3x. Now here the value of x has been given as 3, by putting this value of x in -3x we get:-

-3*3= -9

Thus, the value of derivative $\small \int_{a}^{f}$ = -9

Ans (e) If x=3 and a=2, let us put these values of x and a, in the value calculated in part (f), therefore we get:-

$\small \frac{1}{f}.\frac{1}{a}.1.x-1.x/x^{2}$

=$\small \frac{1}{f}.\frac{1}{2}.1.3-1.3/3^{2}$

=1/f.(1/2*2*3/9)

=1/f.1/3

Ans (f) v₁ = $\small \left ( \frac{\partial }{\partial x}(x,a) \right )x/f(x,a)$

v_{2 =}$\small \left ( \int_{a}^{f} (x,a)\right )a/f(x,a)$

Let us treat a,f as a constant

Now , take the constant out , we get:-

= $\small \frac{1}{f}\frac{\partial }{\partial x}\left ( x/x,a \right )$

= $\small \frac{1}{f}\frac{1}{a}\frac{\partial }{\partial x}\left ( x/x \right )$

= $\small \frac{1}{f}\frac{1}{a}\left ( \frac{\partial }{\partial x}(x)x-\frac{\partial }{\partial x} (x)x\right/x^{2})$

=$\small \frac{1}{f}.\frac{1}{a}.1.x-1.x/x^{2}$